# If the equation of projection is y=ax-bx^2, find initial velocity(u) and angle of projection(theta)?

Jun 25, 2017

Given that the equation of projection is $y = a x - b {x}^{2}$, we are to find initial velocity($u$) and angle of projection($\theta$)

Horizontal component of velocity of projection is $u \cos \theta$

Vertical component of velocity of projection is $u \sin \theta$

Let at the t th sec after its projection from origin its position be represented by the coordinates $\left(x , y\right)$

So $x = u \cos \theta \times t$

$\implies t = \frac{x}{u \cos \theta} \ldots . . \left[1\right]$

Again

$y = u \sin \theta \times t - \frac{1}{2} g \times {t}^{2.} \ldots . \left[2\right]$

Combining [1] and[2] we get

$y = u \sin \theta \times \frac{x}{u \cos \theta} - \frac{1}{2} g \times {x}^{2} / \left({u}^{2} {\cos}^{2} \theta\right)$

$\implies y = x \tan \theta - \frac{g {x}^{2}}{2 {u}^{2} {\cos}^{2} \theta} \ldots . \left[3\right]$

Comparing equation [3] with the given equation we get

$a = \tan \theta$

$\implies \theta = {\tan}^{-} 1 a$

and

$b = \frac{g}{2 {u}^{2} {\cos}^{2} \theta}$

$\implies {u}^{2} = \frac{g}{2 b} {\sec}^{2} \theta$

$\implies {u}^{2} = \frac{g}{2 b} \left(1 + {\tan}^{2} \theta\right)$

$\implies {u}^{2} = \frac{g}{2 b} \left(1 + {a}^{2}\right)$

$\implies u = \sqrt{\frac{g}{2 b} \left(1 + {a}^{2}\right)}$

Jun 25, 2017

See below.

#### Explanation:

This is a typical kinematic problem. Assuming that the movement begins at the referential origin we have

$\left\{\begin{matrix}x = {v}_{x} t \\ y = {v}_{y} t - \frac{1}{2} g {t}^{2}\end{matrix}\right.$

where

$t$ is the elapsed time
${v}_{x}$ the $x$-axis velocity component
${v}_{y}$ the $y$-axis velocity component
$g$ the gravity acceleration.

Now the orbit equation is

$y - a x + b {x}^{2} = 0$ substituting the parametric equations we have

$\frac{1}{2} \left(g t + 2 a {v}_{x} - 2 b t {v}_{x}^{2} - 2 {v}_{y}\right) t = 0$

This condition must be true for all $t$ so

$\left(g t + 2 a {v}_{x} - 2 b t {v}_{x}^{2} - 2 {v}_{y}\right) = 0 , \forall t$

Now grouping coefficients

$\left(g - 2 b {v}_{x}^{2}\right) t + 2 \left(a {v}_{x} - {v}_{y}\right) = 0$ so

$\left\{\begin{matrix}g - 2 b {v}_{x}^{2} = 0 \\ a {v}_{x} - {v}_{y} = 0\end{matrix}\right.$

Now solving for ${v}_{x} , {v}_{y}$

$\left\{\begin{matrix}{v}_{x} = \sqrt{\frac{g}{2 b}} \\ {v}_{y} = a \sqrt{\frac{g}{2 b}}\end{matrix}\right.$

Now $v = \sqrt{{v}_{x}^{2} + {v}_{y}^{2}} = \sqrt{\left(\frac{g}{2}\right) \left(\frac{{a}^{2} + 1}{b}\right)}$

$\theta = \arctan \left({v}_{y} / {v}_{x}\right) = \arctan \left(a\right)$